Advertisements
Advertisements
प्रश्न
A particle is fired vertically upward from earth's surface and it goes up to a maximum height of 6400 km. Find the initial speed of particle.
Advertisements
उत्तर
The particle attained a maximum height 6400 km, which is equal to the radius of the Earth.
Total energy of the particle on the Earth's surface is given by \[E_e = \frac{1}{2} M V^2 + \left( \frac{- gmM}{r} \right) . . . . \left( 1 \right)\]
Now, total energy of the particle at the maximum height is given by
\[E_3 = \left( \frac{- GMm}{R + h} \right) + 0\]
\[ \Rightarrow E_3 = \left( \frac{- GMm}{2R} \right) . . . \left( 2 \right) \left( \therefore g = R \right)\]
\[- \frac{GMm}{R} + \frac{1}{2}m v^2 = - \frac{GMm}{2R}\]
\[ \Rightarrow \left( \frac{1}{2} \right)m v^2 = GMm\left( - \frac{1}{2R} + \frac{1}{R} \right)\]
\[ \Rightarrow v^2 = \frac{GM}{R}\]
\[ = \frac{6 . 67 \times {10}^{- 11} \times 6 \times {10}^{24}}{6400 \times {10}^3}\]
\[ = \frac{40 . 02 + {10}^{13}}{6 . 4 \times {10}^6}\]
\[ = 6 . 2 \times {10}^7 = 0 . 62 \times {10}^8 \]
\[ \therefore v = \sqrt{0 . 62 \times {10}^8}\]
\[ = 0 . 79 \times {10}^4 m/s\]
\[ = 79 km/s\]
APPEARS IN
संबंधित प्रश्न
Is there any meaning of "Weight of the earth"?
The earth revolves round the sun because the sun attracts the earth. The sun also attracts the moon and this force is about twice as large as the attraction of the earth on the moon. Why does the moon not revolve round the sun? Or does it?
The acceleration of the moon just before it strikes the earth in the previous question is
Suppose, the acceleration due to gravity at the earth's surface is 10 m s−2 and at the surface of Mars it is 4⋅0 m s−2. A 60 kg passenger goes from the earth to the Mars in a spaceship moving with a constant velocity. Neglect all other objects in the sky. Which part of the following figure best represents the weight (net gravitational force) of the passenger as a function of time?
If the acceleration due to gravity at the surface of the earth is g, the work done in slowly lifting a body of mass m from the earth's surface to a height R equal to the radius of the earth is
Take the effect of bulging of earth and its rotation in account. Consider the following statements :
(A) There are points outside the earth where the value of g is equal to its value at the equator.
(B) There are points outside the earth where the value of g is equal to its value at the poles.
Find the height over the Earth's surface at which the weight of a body becomes half of its value at the surface.
What is the acceleration due to gravity on the top of Mount Everest? Mount Everest is the highest mountain peak of the world at the height of 8848 m. The value at sea level is 9.80 m s−2.
Find the acceleration due to gravity in a mine of depth 640 m if the value at the surface is 9.800 m s−2. The radius of the earth is 6400 km.
If the acceleration due to gravity becomes 4 times its original value, then escape speed ____________.
Explain the variation of g with latitude.
Explain the variation of g with depth from the Earth’s surface.
If both the mass and the radius of the earth decrease by 1%, then the value of acceleration due to gravity will
The earth is an approximate sphere. If the interior contained matter which is not of the same density everywhere, then on the surface of the earth, the acceleration due to gravity ______.
Which of the following options are correct?
- Acceleration due to gravity decreases with increasing altitude.
- Acceleration due to gravity increases with increasing depth (assume the earth to be a sphere of uniform density).
- Acceleration due to gravity increases with increasing latitude.
- Acceleration due to gravity is independent of the mass of the earth.
If R is the radius of the earth and g is the acceleration due to gravity on the earth's surface, the mean density of the earth is ______.
The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface of the earth will, be ______.
(Radius of earth = 6400 km)
